<< Ïðåäûäóùàÿ ñòð. 75(èç 135 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>
??
? ?2
? ln ? ? + c3 ? 2 + c4 , c = 2.
2 4
For 11? (2.3) we did not find solutions. A particular solution of 12? (2.3) is
c2
?
? = 2c ? ,
12 . f = c, g = 0,
2
?
c2
?c1 e?1 ? + c2 e?2 ? ,
? > ?2 (1 + c),
?
?
? 4
?
c2
h = e?? (c1 + c2 ?), = ?2 (1 + c),
?
? 4
?
?
? ?? c2
?e (c cos ?? + c sin ??), < ?2 (1 + c),
1 2
4
312 W.I. Fushchych, W.M. Shtelen, S.L. Slavutsky

1 + (c/2) ±(c2 /4) ? ?2 (1 + c) 1 + (c/2)
?1,2 = , ?= ,
2(1 + ?2 ) 2(1 + ?2 )
?2 (1 + c) ? (c2 /4)
?= .
2(1 + ?2 )

A particular solution of 13? (2.3) is
1
13? . ? = ? (c2 + c2 ). (2.4)
f = c1 , g = c2 , h = 0,
21 2

Consider system 14? (2.3). The last equation of 14? (2.3) immediately gives

(2.5)
f = c/?

(as before, c is an arbitrary constant). Substituting (2.5) into the remaining equations
of 14? (2.3) we get

d2 2c d
4 2 (?g) + 1 ? (2.6)
(?g) = 0
d? ? d?

and
1
? ?
4? h + (? + 4 ? 2c)h + h = 0. (2.7)
2
Equation (2.6) can be easily integrated and the result is
?
c1 c2
xc/2 e?x/4 dx + (2.8)
g(?) = .
? ?
In particular, when c = 0, the general solution of equation (2.6) takes the form
c1 ??/4 c2
(2.9)
g(?) = e +.
? ?
Equation (2.7) is in itself an equation for a degenerate hypergeometric function and it
can be rewritten in standard Whittaker form

4x2 w ? (x2 ? 4kx + 4m2 ? 1)w = 0, (2.10)
?

where w = w(k, m, x); k, m are parameters, by the substitution
c c?
h(?) = ? (c?2)/4 e??/8 w ,? , (2.11)
.
4 44
When c = 0, the substitution
?
h(?) = e?? Z0 (? ),
? (2.12)
?=
8
reduces (2.7) to the modified Bessel equation of null order, that is

? ?
? ? ?
? Z 0 + Z 0 ? ? Z0 = 0. (2.13)
Reduction and exact solutions of the Navier–Stokes equations 313

Summarizing results (2.5)–(2.12) we can write down the general solution of 14? (2.3)
as follows
?
c c1 c2
xc/2 e?x/4 dx +
?
14 . f= , g= ,
? ? ?
(2.4)
?
c2
c c? 1
(c?2)/4 ??/8
,? , ?=? 2
h=? e w , + g (y)dy + c3 .
4 44 2? 2
(We continue to numerate solutions of reduced NS equations 1? –19? (2.3) as n? (2.4),
where n? = 1? –19? indicates the corresponding ansatz of table 1.) When c = 0 we get
from 14? (2.4) the following particular solution of 14? (2.3)
c1 ??/4 c2
+ , h = e??/8 Z0 (?/8),
14?? . f = 0, ?
g=
e
? ?
(2.4)
c2 ? e?y/2 ? ?y/2
c2 e
?=? 2 1
+ dy + c1 c2 dy + c3 ,
y2 y2
2? 2
?
where Z0 is modified Bessel function satisfying equation (2.12).
Consider system 15? (2.3). The last equation in it gives
c 1
?. (2.13)
f=
?2
The rest equations of 15? (2.3) take the form

d2 c d
(?g) + 1 ? (2.14)
2 (?g) = 0,
d? 2 ? d?
2
c 1
+ g2 ? , (2.15)
2? =
?
? 4

1 c 1
? ?
?+1? h ? h = 0. (2.16)
?h +
2 2 8

Equations (2.14), (2.15) can be easily integrated and the result is as follows
?
c1 c2
xc/2 e?x/2 dx + (2.17)
g= ,
? ?
?
c2
1 1
g 2 (y)dy ? ? ?. (2.18)
?=
2 2? 8

Equation (2.16) is reduced to the Whittaker equation (2.10) by the substitution

c?3 c ?
h(?) = ? (c?2)/4 e??/4 w ,? , (2.19)
.
4 42

Note, when c = 3, function w 0, ? 3 , ? is reduced to the modified Bessel function
42
??3/4 (?/4). The general relation is (Bateman and Erdelyi [2])
Z
v
? (2.20)
w(0, m, x) = xZm (x/2).
314 W.I. Fushchych, W.M. Shtelen, S.L. Slavutsky

So, we can write down the general solution of reduced ns equations 15? (2.3) in the
form
?
c 1 c1 c2
xc/2 e?x/2 dx +
?
f= ? ,
15 . g= ,
?2 ? ?
(2.4)
c?3 c ? ?
c2
1 1
(c?2)/4 ??/4
,? , g (y)dy ? ? ?,
2
h=? e w , ?=
4 42 2 2? 8

where w satisfies the Whittaker equation (2.10).
Consider system 16? (2.3). The two last equations of it give rise to

c?
(2.21)
h = c, ?= + c1 .
2

Taking into account (2.21) we can rewrite the rest equations of system 16? (2.3) as
follows

1 1
? ? ? c f? + f = 0, (2.22)
f+
2 2

1 1
? ? c g + g = 0. (2.23)
g+
? ?
2 2

By substituting

1
??c (2.24)
f (?) = F (? ), ?=
2

into (2.22), we obtain the following equation:

d2 F dF
(2.25)
+ 2? + 2F = 0.
2
d? d?

The general solution of (2.25) is
?
?? 2 2
ey dy . (2.26)
F (? ) = e c2 + c3

Summarizing results (2.21)–(2.26) we write down the general solution of equations
16? (2.3):

(?/2)?c
2
? 2
?
f = exp ? ?c ey dy ,
16 . c2 + c3
2
(?/2)?c
2
? (2.4)
2
g = exp ? ?c ey dy ,
c4 + c5
2
c?
h = c, ? = + c1 .
2
Reduction and exact solutions of the Navier–Stokes equations 315

In the same way we find solutions of reduced equations 17? –19? (2.3). The soluti-
ons are as follows
 << Ïðåäûäóùàÿ ñòð. 75(èç 135 ñòð.)ÎÃËÀÂËÅÍÈÅ Ñëåäóþùàÿ >>